My Math Forum Summation Problem

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 April 30th, 2017, 03:39 AM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 211 Thanks: 2 Summation Problem Hello Guys! Can someone help me with this! How many pair of positive integers of $(m,n)$ are there satisfying $\sum_{i=1}^{n} i! = m!$ I guess it has only one pair of solution : $(m,n) =(1,1)$ Then starting and ending value of Summation should be equal as I start at $i=1$ and we need $n$ to be 1. Will it make sense or Is it wrong to have both values same ? Please let me know Thank you
 April 30th, 2017, 05:00 AM #2 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 For $\ \ \ \ n > 1 \ \ \ \$ , $\ \ \ \ n < m \ \ \ \$ otherwise equality has no chance. Is this obvious to you? Suppose $\ \ \ \ n = m - 1$ We need to determine if $\ \ \ \ \sum_{i=1}^{m-1}i! \ \ \ \$ can catch up to $\ \ \ \ m!$ Let's see how the desired equality would look $1! + 2! + 3! + ... + (m-1)! = 1 \times 2 \times 3 \times ... \times (m-1) \times m$ $1! + 2! + 3! + ... + (m-1)! = (m-1)! \times m$ $1! + 2! + ... + (m-1)! = (m-1)! + (m-1)! + ... + (m-1)!$ The left hand side has $\ \ \ \ m-1 \ \ \ \$ sums , the right hand side has $\ \ \ \ m \ \ \ \$ sums. Moreover , every sum on the left except the last sum is smaller than any sum on the right. So no chance to catch up Last edited by agentredlum; April 30th, 2017 at 05:06 AM.
 April 30th, 2017, 05:29 AM #3 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 You can have both values the same $\sum_{i = 1}^{1} f(i) = f(1)$
April 30th, 2017, 05:35 AM   #4
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Quote:
 Originally Posted by agentredlum You can have both values the same $\sum_{i = 1}^{1} f(i) = f(1)$
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